Free CAS MAS-I (Modern Actuarial Statistics I) Statistics Practice Questions
Statistical inference on CAS Exam MAS-I covers hypothesis testing, confidence interval construction, Bayesian estimation methods, Monte Carlo simulation, and applied statistics for P&C actuarial problems (CAS).
182 questions55 easy70 medium57 hard2026 syllabus
Sample Questions
Question 1
Easy
An actuary records five independent claim severities (in thousands):
8,12,15,18,22
Calculate the unbiased sample variance.
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Correct Answer: E
Solution
E is correct. First compute the sample mean: Xˉ=58+12+15+18+22=575=15. The deviations from the mean are −7,−3,0,3,7, with squared deviations 49,9,0,9,49, summing to 116. The unbiased sample variance uses the divisor n−1=4: S2=n−1∑(Xi−Xˉ)2=4116=29. The value 29 falls in the interval [25,30), so the answer is E.
Question 2
Medium
Let X1,X2,X3,X4,X5 be a random sample of five independent observations from a continuous Uniform(0,10) distribution. Calculate the probability that the sample maximum X(5) exceeds 8.
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Correct Answer: B
Solution
B is correct. The maximum X(5) is at most 8 if and only if every observation is at most 8. By independence, P(X(5)≤8)=i=1∏5P(Xi≤8)=[FX(8)]5=(8/10)5=0.85=0.32768. Therefore P(X(5)>8)=1−0.32768=0.67232. This value falls in the interval [0.65,0.80).
Question 3
Hard
An actuary observes six iid claim amounts drawn from an exponential distribution with mean θ=250. Let X(1)≤X(2)≤⋯≤X(6) denote the order statistics.
Calculate P(X(3)>200).
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Correct Answer: B
Solution
B is correct.
Step 1: Translate the event in terms of how many observations fall on each side of 200. X(3)>200 means the third-smallest of the six exceeds 200, equivalently at most two of the six observations are ≤200.
Step 2: Define a binomial counter. Let Y=#{i:Xi≤200}. Since the Xi are iid, Y∼Binomial(6,p) with p=P(X≤200).
Step 3: Compute p using the exponential cdf. p=1−e−200/250=1−e−0.8≈1−0.4493=0.5507,1−p≈0.4493.
Step 4: Sum the binomial probabilities for Y≤2. P(Y=0)=(0.4493)6≈0.00823 P(Y=1)=6(0.5507)(0.4493)5≈6⋅0.5507⋅0.01831≈0.06051 P(Y=2)=15(0.5507)2(0.4493)4≈15⋅0.30327⋅0.04075≈0.18539.
The single-observation tail P(X>200)=e−0.8≈0.449 is a tempting shortcut but ignores the order-statistic structure entirely; it answers a different question.
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